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Mathematics 14 Online
OpenStudy (anonymous):

Graph X^2 + (y +3)^2=9

OpenStudy (accessdenied):

Are you familiar with what sort of shape you'll get from an equation of this form?

OpenStudy (anonymous):

Not this one

OpenStudy (accessdenied):

Okay. Equations of the form \((x - h)^2 + (y - k )^2 = r^2\) are considered circles. The center of the circle is (h, k), and the radius is r. So, you can usually make comparisons between the 'standard form'and the given equation to figure out your center + radius, and graph from that. :)

OpenStudy (anonymous):

will the center b (0,-3)?

OpenStudy (accessdenied):

Yes, that is correct.

OpenStudy (anonymous):

So then this will be a circle intersecting at ±3 all round

OpenStudy (accessdenied):

Hmm... well, the circle simply has a radius of three, so basically all the points that are a distance of 3 from (0, -3) are points on your graph. It makes sense if you consider r the distance between the center and all the points on the circle that are 3 units away. :)

OpenStudy (anonymous):

|dw:1352106893652:dw|

OpenStudy (accessdenied):

The exact center of the circle will be at (0, -3), like this: |dw:1352107269477:dw|

OpenStudy (anonymous):

Since the radius is 9 and the √9=3

OpenStudy (accessdenied):

Our radius is determined as r^2 = 9 => r = 3. :) So, we can draw in all the points around (0, -3) that are a distance of 3 units away for the circle.

OpenStudy (accessdenied):

|dw:1352107502707:dw| So, we can pick out a few simple ones that line up on the axes, and then we have to estimate from there. My graph isn't drawn particularly well, so it doesn't look very nice, but you should be able to get a circle. :P

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